Problem: What is the slope of the line through the points $(-5, -2)$ and $(-2, -6)$ ? {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9} {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9}
Picture an airplane moving from left to right. If the airplane is taking off ${/}$ , the slope is positive. If it's landing ${\backslash}$ , the slope is negative. When the plane has reached cruising altitude ${-}$ , the slope is 0. ${\text{Airplane Taking Off}}$ ${\text{Airplane Landing}}$ The quicker the airplane takes off, the steeper the positive slope, which means the answer will be a higher number than when the airplane takes off slowly. The quicker the airplane lands, the steeper the negative slope, which means the answer will be much more negative than when it lands slowly. The equation for the slope is $m = \dfrac{{y_2} - {y_1}}{{x_2} - {x_1}}$ for points $({-5}, {-2})$ and $({-2}, {-6})$ Substituting in, we get $m = \dfrac{{-6} - {(-2)}}{{-2} - {(-5)}} = \dfrac{{-4}}{{3}}$ So, the slope $m$ is $-\dfrac{4}{3}$.